Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in CPP):

Solution:
#include <cstdio>

const int maxn=3005;

double dp[maxn][maxn];
double l[maxn];

int main(){
	int n;
	scanf("%d", &n);
	for(int i=0; i<n; i++){
		scanf("%lf", &l[i]);
	}
	dp[0][0]=1;
	for(int i=1; i<=n; i++){
		dp[i][0]=dp[i-1][0]*(1.0-l[i-1]);
		for(int j=1; j<=i; j++){
			dp[i][j]=dp[i-1][j-1]*l[i-1]+dp[i-1][j]*(1.0-l[i-1]);
		}
	}
	double sol=0;
	for(int i=n/2+1; i<=n; i++){
		sol+=dp[n][i];
	}
	printf("%.9lf\n", sol);
	return 0;
}